How do you solve the following system: $4 x + 3 y = - 7, y + 4 x = 16$ $4x+3y= -7, y + 4x = 16$ ?

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Answer 1 · 2016-02-14T16:46:26

The solution for the system of equations is:
$x = \frac{55}{8}$ $color(blue)(x=55/8$

$y = - \frac{23}{2}$ $color(blue)(y=-23/2$

$4 x + 3 y = - 7$ $color(blue)(4x)+3y=-7$ ..........equation $1$ $1$

$4 x + y = 16$ $color(blue)(4x)+y=16$ .................equation $2$ $2$

Solving by elimination

Subtracting equation $2$ $2$ from $1$ $1$
$cancelcolor(blue)(4x)+3y=-7$

$-cancelcolor(blue)(4x)-y=-16$

$2y=-23$

$color(blue)(y=-23/2$

Finding $x$ from equation $1$

$4x+3y=-7$

$4x+3xx(-23/2)=-7$

$4x-69/2=-7$

$4x=-7 +69/2$

$4x=-14/2 +69/2$

$4x=55/2$

$x=55/(2xx4)$

$color(blue)(x=55/8$

How do you solve the following system: 4x+3y= -7, y + 4x = 16 4x+3y=−7,y+4x=164x+3y= -7, y + 4x = 16 ?